#### Answer

1) The interval is $(-2.1213,-1.8708)$
2) $\delta=0.1213$

#### Work Step by Step

Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-(-2)|\lt\delta\Rightarrow|(x^2)-4|\lt0.5$$
1) Find the interval around $-2$ on which $|(x^2)-4|\lt0.5$ holds.
Solve the inequality: $$|(x^2)-4|\lt0.5$$ $$-0.5\lt x^2-4\lt0.5$$ $$3.5\lt x^2\lt4.5$$
Now we need to take the square root. However, there are 2 cases, whether $x\ge0$ or $x\lt0$.
- For $x\ge0$: $1.8708\lt x\lt2.1213$
- For $x\lt0$: $-2.1213\lt x\lt-1.8708$
Yet, here we are looking for the interval around $-2$, which means $x\lt0$.
Therefore, we choose only the open interval around $-2$, which is $(-2.1213,-1.8708)$.
2) Give a value for $\delta$
- From $-2.1213$ to $-2$: $|-2.1213|-|-2|=0.1213$
- From $-1.8708$ to $-2$: $|-2|-|-1.8708|=0.1292$
The nearer endpoint to $-2$ is $-2.1213$, and the distance between them is $0.1213$.
So if we take $\delta=0.1213$ or any smaller positive number, then $0\lt|x-(-2)|\lt0.1213$, meaning all $x$ would be placed in the interval $(-2.1213,-1.8708)$ so that $|(x^2)-4|\lt0.5$.
In other words, $$0\lt |x-(-2)|\lt0.1213\Rightarrow|(x^2)-4|\lt0.5$$